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3 years ago

What are the solutions of the equation c

Mathematics

1 answer:

kondaur [170]3 years ago

5 0

Answer:

Be more detailed, BC i can only give you a general answer a^2 + b^2 = c^2

Step-by-step explanation:

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During a volleyball training session, Keela drank twice as much water as Carol, and Xavier drank 100 mL more water than Keela. B

vladimir2022 [97]

Answer:

580ml

Step-by-step explanation:

Let The quantity of water drank by Carol be x

Now given that Keela drank twice as much water as Carol which means the quantity of water drank by Keela is 2x

Also given that Xavier drank 100 mL more water than Keela which means the quantity of water drank by Xavier is 2x+100

Total water consumed by all the three is 3 litres=3000 ml

Total water drank = Water drank by Keela + Water drank by Xavier +Water drank by carol

3000=x+2x+2x+100

5x=2900

x=580ml

Carol drank 580ml of water.

5 0

3 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

4 0

3 years ago

Determine the value of a + b in the system of equations below. 5a + 4b = 22 7a-b= 11

dexar [7]

Answer:

We have a + b as;

2 + 3 = 5

Step-by-step explanation:

Here, we want go get the value of a and b

from the second equation, we can get an expression for b

We have this as;

7a - b = 11

Thus,

b = 7a - 11

Now, from here, we can substitute the value of b into the first equation

We have this as;

5a + 4(7a - 11) = 22

5a + 28a -44 = 22

33a = 22 + 44

33a = 66

a = 66/33

a = 2

Recall;

b = 7a - 11

substituting the value of a from above;

b = 7(2) - 11

b = 14 - 11

b = 3

6 0

3 years ago

PLS HELP IM STUCK ON THIS ONE PROBLEM!! PLEASE SHOW STEPS

natka813 [3]

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3 0

3 years ago

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Natalija [7]

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8 0

3 years ago

What are the solutions of the equation c​

What are the solutions of the equation c